A circular arc of radius r carrying current ' $I$ ' subtends an angle $\frac{\pi}{8}$ at its entre. The…
A circular arc of radius r carrying current ' $I$ ' subtends an angle $\frac{\pi}{8}$ at its entre. The radius of a metal wire is uniform. The magnetic induction at the centre of circular arc is ( $\mu_0=$ permeability of free space)
$\frac{\mu_0 \mathrm{I}}{8 \mathrm{r}}$
$\frac{\mu_0 I}{32 r}$
$\frac{\mu_0 \mathrm{I}}{64 \mathrm{r}}$
$\frac{\mu_0 \mathrm{I}}{16 \mathrm{r}}$
Solution
The magnetic induction at the centre of circular arc is
$B=\frac{\mu_0 I}{4 \pi r} \times \theta$
$\therefore \quad B=\frac{\mu_0 \dot{I}}{4 \pi r} \times \frac{\pi}{8} \quad \cdots\left[\because \theta=\frac{\pi}{8}\right]$
$\therefore \quad B=\frac{\mu_0 I}{32 r}$